Question: The equation of a circle $C$ is $x^2+y^2+8x-2y+1 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+8x) + (y^2-2y) = -1$ $(x^2+8x+16) + (y^2-2y+1) = -1 + 16 + 1$ $(x+4)^{2} + (y-1)^{2} = 16 = 4^2$ Thus, $(h, k) = (-4, 1)$ and $r = 4$.